12th Grade > Mathematics
TRIGONOMETRIC FUNCTIONS MCQs
Total Questions : 30
| Page 2 of 3 pages
Answer: Option C. -> 2a23
:
C
Adding and subtracting the given relation,
we get (m+n)=acos3α+3acosαsin2α
+ 3acos2α.sinα+asin3α
= a(cosα+sinα)3
and similarly (m−n)=a(cosα−sinα)3
Thus, (m+n)23+(m−n)23
=a23 {(cosα+sinα)2+(cosα−sinα)2}
=a23 [2(cos2α+sin2α)]= 2a23.
:
C
Adding and subtracting the given relation,
we get (m+n)=acos3α+3acosαsin2α
+ 3acos2α.sinα+asin3α
= a(cosα+sinα)3
and similarly (m−n)=a(cosα−sinα)3
Thus, (m+n)23+(m−n)23
=a23 {(cosα+sinα)2+(cosα−sinα)2}
=a23 [2(cos2α+sin2α)]= 2a23.
Answer: Option C. -> cotα
:
C
=tanα+2tan2α+4tan4α+8cot8α
tanα+2tan2α+4[sin4αcos4α+2cos8αsin8α]
=tanα+2tan2α+4[cos4αcos8α+sin4αsin8α+cos4αcos8αsin8αcos4α]
=tanα+2tan2α+4[cos4α+cos4αcos8αsin8αcos4α]
=tanα+2tan2α+4[cos4α(1+cos8α)cos4αsin8α]
=tanα+2tan2α+4[2cos24α2sin4αcos4α]
=tanα+2tan2α+4cot4α
=tanα+2(tan2α+2cot4α)
=tanα+2[sin2αcos2α+2cos4αsin4α]
=tanα+2[cos2α(1+cos4α)sin4αcos2α]
=tanα+2cot2α=sinαcosα+2cos2αsin2α
=cosα+cosαcos2αsin2αcosα
=1+cos2αsin2α=2cos2α2sinαcosα=cotα
:
C
=tanα+2tan2α+4tan4α+8cot8α
tanα+2tan2α+4[sin4αcos4α+2cos8αsin8α]
=tanα+2tan2α+4[cos4αcos8α+sin4αsin8α+cos4αcos8αsin8αcos4α]
=tanα+2tan2α+4[cos4α+cos4αcos8αsin8αcos4α]
=tanα+2tan2α+4[cos4α(1+cos8α)cos4αsin8α]
=tanα+2tan2α+4[2cos24α2sin4αcos4α]
=tanα+2tan2α+4cot4α
=tanα+2(tan2α+2cot4α)
=tanα+2[sin2αcos2α+2cos4αsin4α]
=tanα+2[cos2α(1+cos4α)sin4αcos2α]
=tanα+2cot2α=sinαcosα+2cos2αsin2α
=cosα+cosαcos2αsin2αcosα
=1+cos2αsin2α=2cos2α2sinαcosα=cotα
Answer: Option D. -> - 1665
:
D
We have sin A = 45 and cos B = - 1213
Now, cos(A + B) = cos A cos B - sin A sin B
= √1−1625(−1213) - 45(−√1−144169)
= - 35× 1213 - 45(−513) = - 1665
(Since A lies in first quadrant and B lies in third quadrant).
:
D
We have sin A = 45 and cos B = - 1213
Now, cos(A + B) = cos A cos B - sin A sin B
= √1−1625(−1213) - 45(−√1−144169)
= - 35× 1213 - 45(−513) = - 1665
(Since A lies in first quadrant and B lies in third quadrant).
Answer: Option B. -> √32
:
B
cot215∘−1cot215∘+1 = cos215∘sin215∘−1cos215∘sin215∘+1
= cos215∘−sin215∘cos215∘+sin215∘ = cos(30∘) = √32
:
B
cot215∘−1cot215∘+1 = cos215∘sin215∘−1cos215∘sin215∘+1
= cos215∘−sin215∘cos215∘+sin215∘ = cos(30∘) = √32
Answer: Option D. -> -5
:
D
Minimum value of (3sinθ+4cosθ) is -√32+42 i.e., -5.
:
D
Minimum value of (3sinθ+4cosθ) is -√32+42 i.e., -5.
Answer: Option C. -> −1−cotα
:
C
√cosec2α+2cotα=√1+cot2α+2cotα=|1+cotα|
But 3π4<α<π⇒cotα<−1⇒1+cotα<0
Hence, |1+cotα|=−(1+cotα)
:
C
√cosec2α+2cotα=√1+cot2α+2cotα=|1+cotα|
But 3π4<α<π⇒cotα<−1⇒1+cotα<0
Hence, |1+cotα|=−(1+cotα)
Answer: Option C. -> a2+b2
:
C
We have sin(α−β) = sin(θ−β−¯¯¯¯¯¯¯¯¯¯¯¯¯θ−α)
= sin(θ−β)cos(θ−α)−cos(θ−β)sin(θ−α)
= ba - √1−b2√1−a2
And cos(α−β)=cos(θ−β−¯¯¯¯¯¯¯¯¯¯¯¯¯θ−α)
= cos(θ−β)cos(θ−α)+sin(θ−β)sin(θ−α)
= a√1−b2+b√1−a2
∴ Given expression is cos2(α−β)+2absin(α−β)
= (a√1−b2+b√1−a2)2 + 2ab{ab−√1−a2√1−b2}
= a2+b2.
Trick:Put α=30∘, β=60∘ and θ=90∘,
then a = 12, b = 12
∴ cos2(α−β)+2absin(α−β) = 34 + 12×(- 12) = 12
which is given by option (c).
:
C
We have sin(α−β) = sin(θ−β−¯¯¯¯¯¯¯¯¯¯¯¯¯θ−α)
= sin(θ−β)cos(θ−α)−cos(θ−β)sin(θ−α)
= ba - √1−b2√1−a2
And cos(α−β)=cos(θ−β−¯¯¯¯¯¯¯¯¯¯¯¯¯θ−α)
= cos(θ−β)cos(θ−α)+sin(θ−β)sin(θ−α)
= a√1−b2+b√1−a2
∴ Given expression is cos2(α−β)+2absin(α−β)
= (a√1−b2+b√1−a2)2 + 2ab{ab−√1−a2√1−b2}
= a2+b2.
Trick:Put α=30∘, β=60∘ and θ=90∘,
then a = 12, b = 12
∴ cos2(α−β)+2absin(α−β) = 34 + 12×(- 12) = 12
which is given by option (c).
Answer: Option B. -> 34
:
B
Since cos6α+sin6α+Ksin22α=1 using formula a3+b3=(a+b)3−3ab(a+b) and on solving, we get the required result i.e., K=34
:
B
Since cos6α+sin6α+Ksin22α=1 using formula a3+b3=(a+b)3−3ab(a+b) and on solving, we get the required result i.e., K=34
Answer: Option B. -> tan (A−B2)
:
B
We have sin A = n sin B ⇒ n1 = sinAsinB
⇒ n−1n+1 sinA−sinBsinA+sinB = 2cosA+B2sinA−B22sinA+B2cosA−B2
= tan A−B2 cot A+B2
⇒ n−1n+1 tan(A+B2) = tan A−B2.
:
B
We have sin A = n sin B ⇒ n1 = sinAsinB
⇒ n−1n+1 sinA−sinBsinA+sinB = 2cosA+B2sinA−B22sinA+B2cosA−B2
= tan A−B2 cot A+B2
⇒ n−1n+1 tan(A+B2) = tan A−B2.
Answer: Option B. -> 75√2
:
B
We have cos θ =
35 and cos ϕ =
45.
Therefore cos(θ−ϕ)=cosθcosϕ+sinθsinϕ
=
35 .
45 +
45 .
35 =
2425
But 2cos2(θ−ϕ2)=1+cos(θ−ϕ) = 1 +
2425 =
4950
∴ cos2(θ−ϕ2) =
4950. Hence, cos(θ−ϕ2) =
75√2.
:
B
We have cos θ =
35 and cos ϕ =
45.
Therefore cos(θ−ϕ)=cosθcosϕ+sinθsinϕ
=
35 .
45 +
45 .
35 =
2425
But 2cos2(θ−ϕ2)=1+cos(θ−ϕ) = 1 +
2425 =
4950
∴ cos2(θ−ϕ2) =
4950. Hence, cos(θ−ϕ2) =
75√2.
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